Question

Evaluate:
$\rm \displaystyle \lim_{x \to \infty}\ \frac{(2x-3)^{2}(3x+2)^{3}}{(2x+1)^{5}}$

Answer 1

To evaluate those limits having x goes to infinite.look
$\lim_{x \to \infty}\ \frac{(2x-3)^{2}(3x+2)^{3}}{(2x+1)^{5}} \\ \\ \lim_{x \to \infty}\ \frac{ {x}^{2} (2- \frac{3}{x} )^{2} {x}^{3} (3+ \frac{2}{x} )^{3}}{ {x}^{5} (2+ \frac{1}{x} )^{5}} \\ \\ \lim_{x \to \infty}\ \frac{ {x}^{5} (2- \frac{3}{x} )^{2} (3+ \frac{2}{x} )^{3}}{ {x}^{5} (2+ \frac{1}{x} )^{5}} \\ \\ \lim_{x \to \infty}\ \frac{ (2- \frac{3}{x} )^{2} (3+ \frac{2}{x} )^{3}}{ (2+ \frac{1}{x} )^{5}} \\ \\ apply \: limit \\ \\ \frac{ (2- \frac{3}{ \infty } )^{2} (3+ \frac{2}{ \infty } )^{3}}{ (2+ \frac{1}{ \infty } )^{5}} \\ \\ \\ = \frac{ (2- 0 )^{2} (3 + 0 )^{3}}{ (2+ 0 )^{5}} \\ \\ = \frac{4 \times 27}{32} \\ \\ \lim_{x \to \infty}\ \frac{(2x-3)^{2}(3x+2)^{3}}{(2x+1)^{5}} = \frac{27}{8}$
Hope it helps.